Complex Numbers - Bilkent University

1 Numbers are Numbers of the forma+bi, whereaandbare real, andwhereiis an element withi2= 1. Complex Numbers , just as the real Numbers ,form a field: (a+bi) + (c+di) = (a+c) + (b+d)i, (a+bi)(c+di) = (ac bd) + (ad+bc)i; a+bic+di=(a+bi)(c di)(c+di)(c di)=ac+bdc2+d2+bc adc2+d2iwheneverc+di6= elementa+bi=a biis called the conjugate ofa+bi, the map itself is calledcomplex :(1)xy=xy;(2)x=x;(3)x=xif and only ifxis real;(4)x= xif and only ifxis imaginary, , ifx=bifor some just as there are real vector spaces (sets of elements that can be added andmultiplied by real Numbers ) there are Complex vector spaces (sets of elements thatcan be added and multiplied by Complex Numbers ). Consider {(xy):x, y C}.

2 This is a 2-dimensionalC- vector space with basis{(10),(01)}since(xy)=x(10)+y(01)for x, y also a real vector space , since we can writex=a+biandy=c+di, and then haveC2={a(10)+b(i0)+c(01)+d(0i):a, b, c, d R}. This means dimCC2= 2 and dimRC2= , polynomials of degree 2 with coefficients inCform aC- vector spaceof dimension 3, since{1, x, x2}is a theory of linear equations with Complex coefficients does not differ at allfrom what you know from the real case. The difference between real and complexvector spaces starts with inner course we can define an inner product onC2by simply puttingu v=uTv(thematrix product of the transpose ofuandv), that is, via(xy) (rs)= (x, y)(rs)=xr+ys.

3 The problem is that(i0)(i0)=i2+ 0 = 1: the dot product of a vectorwith itself can be negative, so we cannot define lengths via||u||= (u, u) (at leastwe cannot define lengths as real Numbers ). Similarly, the vector (1i)is orthogonalto itself with respect to this (naive) definition of an inner correct way to avoid this is defining the inner product differently. In fact,we put (u, v) =uTv. Thus we have ((i0),(i0)) = ( i,0)(i0)= 1, and(i0)haslength inner product onC2has the following properties:(1) (u, u) 0 for allu C2, with equality if and only ifu= 0;(2) (u+u , v) = (u, v) + (u , v);(3) (u, v) = (u, v) for all C;(4) ( u, v) = (u, v);(5) (u, v) =(v, u).The proofs are easy and left as an exercise (I did all of them in class).

4 In a general Complex vector spaceV(such as those occurring in quantum me-chanics), we say that a map ( , ) sending a pair of vectorsu, vto a complexnumber (u, v) is an inner product if it has the properties above. The standard dot12product onCndefined via (u, v) =uTvis an inner product, the naive dot product(u, v) =uTvis can also define a dot product on function spaces such as the vector spaceVof all polynomials with Complex coefficients by putting(p, q) = 10p(x)q(x) example,(x+i,3x 1) = 10(x i)(3x 1)dx= 10(3x2 (1 + 3i)x+i)dx=x3 12(1 + 3i)x2+ix 10= 1 12(1 + 3i) +i=1 i2and(x+i, x+i) = 10(x i)(x+i)dx= 10(x2+ 1)dx= the properties (p+r, q) = (p, q) + (r, q) forp, q, r V; (cp, q) =c(p, q) for allc C; (p, q) =(q, p)is a purely formal exercise (do it!)

5 It remains to see why (p, p) 0: we have(p, p) = 10p(x)p(x) function below the integral is a real valued nonnegative function becausep(x)p(x) 0 for all values ofx. Thus (p, p) 0, with equality if and only ifp= that we have decent inner products, we can define lengths of vectors incomplex inner product spaces via||u||= (u, u) as before. Moreover, the proofof the Gram-Schmidt process carries over word for word, which means that everybasis of a vector space can be transformed into an orthonormal basis using the sameformulas as in the real , ifVis a subspace of a Complex inner product spaceU, then theorthogonal complementV ={u U: (u, v) = 0 for allv V}is a vector space ,and we haveU=V V as well as dimU= dimV+ dimV.